Prove the following equation for the angle between two vectors.

Question

I'm currently reviewing coordinate and matrix transformations for a physics class. In one of the questions, I am asked to prove the following two equations.

Equation 1 \begin{equation} cos^2(\alpha)+cos^2(\beta)+cos^2(\gamma)=1 \end{equation} Equation 2 $$ cos(\theta)=cos(\alpha)cos(\alpha')+cos(\beta)cos(\beta')+cos(\gamma)cos(\gamma') $$

I've proven Eq. 1 using the law of cosines, with help from a separate post here on this site. I have a hunch that the proof for the second equation will be similar to the first, but I'm lost on where I should start. I've explored the law of cosines with two vectors, but I am either missing something, or it's getting me nowhere.

I'm looking for a push in the right direction.

Thank you!

EDIT: Alpha is the angle from the x-axis to the vector, beta is the angle from the y-axis to the vector, and gamma is the angle from the z-axis to the vector.

Answer 1

The proof to Eq. 1 can be found here

For Eq. 2:

The dot product of two vectors is defined to be $$ \mathbf V \cdot \mathbf W =||\mathbf V|| \cdot ||\mathbf W|| cos(\theta) $$ Where $$\mathbf V =(x,y,z), \mathbf W=(x',y',z') $$ We see that $$ cos(\theta) = \frac {\mathbf V \cdot \mathbf W }{||\mathbf V|| \cdot ||\mathbf W||} = \frac{(x,y,z)\cdot(x',y',z')}{(x^2+y^2+z^2)^\frac{1}{2}(x'^2+y'^2+z'^2)^\frac{1}{2}} $$

From the proof for Eq. 1, we see that $$ cos(\alpha)=x(x^2+y^2+z^2)^\frac{-1}{2} $$ $$ cos(\beta)=y(x^2+y^2+z^2)^\frac{-1}{2} $$ $$ cos(\gamma)=z(x^2+y^2+z^2)^\frac{-1}{2} $$

We can then assume that the same relationship holds true for the primed system. i.e., $$ cos(\alpha')=x'(x'^2+y'^2+z'^2)^\frac{-1}{2} $$ $$ cos(\beta')=y'(x'^2+y'^2+z'^2)^\frac{-1}{2} $$ $$ cos(\gamma')=z'(x'^2+y'^2+z'^2)^\frac{-1}{2} $$

By expanding the equation for $cos(\theta)$ above and substituting the values above, we then prove that $$ cos(\theta)=cos(\alpha)cos(\alpha')+cos(\beta)cos(\beta')+cos(\gamma)cos(\gamma') $$

Answer 2

Let u and v be two unit vectors, with direction angles $\alpha, \beta, \gamma$ and $\alpha^{\prime}, \beta^{\prime}, \gamma^{\prime}$, and let $\theta$ be the angle between $u$ and $v$.

Then $\displaystyle \cos\theta=\frac{u\cdot v}{|u||v|}=\frac{\cos(\alpha)\cos(\alpha^{\prime})+\cos(\beta)\cos(\beta^{\prime})+\cos(\gamma)\cos(\gamma^{\prime})}{1\cdot1}$.